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1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
\(\left\{{}\begin{matrix}u=x^2\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2sin2x|^{\pi}_0-\int\limits^{\pi}_0x.sin2xdx\)
\(I=\int\limits^{\dfrac{\pi}{2}}_0\left(1+cosx+x.cosx\right)e^{sinx}dx=\int\limits^{\dfrac{\pi}{2}}_0e^{sinx}dx+\int\limits^{\dfrac{\pi}{2}}_0\left(x+1\right).cosx.e^{sinx}dx=I_1+I_2\)
Xét \(I_2\), đặt \(\left\{{}\begin{matrix}u=x+1\\dv=cosx.e^{sinx}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^{sinx}\end{matrix}\right.\)
\(\Rightarrow I_2=\left(x+1\right).e^{sinx}|^{\dfrac{\pi}{2}}_0-\int\limits^{\dfrac{\pi}{2}}_0e^{sinx}dx=\left(\dfrac{\pi}{2}+1\right)e-1-I_1\)
\(\Rightarrow I=I_1+\left(\dfrac{\pi}{2}+1\right)e-1-I_1=\left(\dfrac{\pi}{2}+1\right)e-1\)
Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)
Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)
Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)
Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)
\(I=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos^2xdx=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\left(\frac{1+\cos2x}{2}\right)dx=\int\limits^{\frac{\pi}{2}}_0\left(x-\frac{1}{2}\right)dx+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos2xdx\)
\(=\left(\frac{1}{2}x^2-\frac{1}{2}x\right)|^{\frac{\pi}{2}}_0+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)d\left(\sin2x\right)=\frac{\pi^2}{8}-\frac{\pi}{4}+\frac{1}{2}\left[\left(2x-1\right)\sin2x|^{\frac{\pi}{2}}_0-\int\limits^{^{\frac{\pi}{2}}_0}_0\sin2x.2dx\right]\)
\(=\frac{\pi^2}{8}-\frac{\pi}{4}+\left(0+\cos2x|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2}{8}-\frac{\pi}{4}-1\)
\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)
Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)
Với \(x=0\Rightarrow t=1\)
Với \(x=\frac{\pi}{2}\Rightarrow t=0\)
\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)
\(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)
\(I=-\frac{1}{2}\int_0^{\frac{\pi}{4}}\left(x^2-4x+3\right)d\cos2x\)
\(=-\frac{1}{2}\left[\left(x^2-4x+3\right)\cos2x\right]_0^{\frac{\pi}{4}}-\int^{^{\frac{\pi}{4}}}_0\cos2xd\left(x^2-4x+\right)\)
\(=\frac{3}{2}+\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\cos2xd=\frac{3}{2}+\frac{1}{2}\int^{^{\frac{\pi}{4}}}_0\left(x-2\right)\sin2x\)
\(=\frac{3}{2}+\frac{1}{2}\left[\left(x-2\right)\sin2x_0^{\frac{\pi}{4}}-\int^4_0\sin2dx\left(x-2\right)\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2+\frac{1}{2}\cos2x_0^{\frac{\pi}{4}}\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\frac{\pi}{4}-2-\frac{1}{2}\right]=\frac{\pi}{8}+\frac{1}{4}\)